Main Puzzle 8M85 by Richard Pavlicek
Professor! You wont believe this monster I picked up with Grover last night. My right-hand opponent opened 3 , and I almost fell out of my chair. Look at this mitt! A cold grand in my own hand!
| A K Q J 10|
A K Q J 9 8 7 6
Nice hand, Timothy, but its hardly cold for seven. Five hearts in one hand, or an opening spade ruff would put you down. In fact, there are only four hands out of 635+ billion that are cold for seven in a suit. It takes a flush hand; anything less, and something could go wrong.
I suppose, but who worries about that. I thought about bidding it deceptively, but nothing was safe with Grover (he even passes cue-bids) so I jumped to 7 . It was cold but a tie for bottom, since everyone else did the same, and five tables were doubled and redoubled. My left-hand opponent smartly passed with two aces. I guess I had to hope for one of those fluke distributions you say to get a tie for top. Strange game, duplicate. At least I was cold for a small slam.
Sorry to disillusion you, Timothy, but you cant even claim six. However unlikely, though consistent with my boson deflection theory, 6 could be set by a spade ruff with a singleton trump, then a diamond to promote a trump trick for left-hand opponent who is also void in diamonds. You needed the 10 to be cold for six.
Whatever. Are you suggesting I shouldnt even bid a slam?
No, but you have to gauge your bidding to bosonic theory. A successful contract is not a simple wave but a complex collision of subnuclear particles. Slam bidding especially requires a strategic approach, not a surge to the verge. Off hand, I dont recall the exact number of hands cold for six of a suit, but its fewer than those cold for 6 NT. Directly confronting this threshold is to deny the mere existence of bosons and could set bridge theory back 30 years.
Professor? Get a life! Every time I bring up a bridge hand you turn it into a lecture on some bozo theory of yours. Does anybody really care how many hands there are?
[Richard raises hand]. In case theres anybody else out there, this will be the June puzzle:
|How many bridge hands are guaranteed to win 12 tricks in a suit?|
For the sake of argument, do not count the four hands cold for 13 tricks. As a tiebreaker, include your best guess to the related question: How many hands are guaranteed to win 12 tricks (but not 13) in notrump? Guaranteed means you must succeed against any distribution, any defense, and without having to guess anything about the layout.
Make your best guesses and click Verify to find out what, if anything, is wrong. Use the help provided to make corrections and repeat. See how many tries it takes you to find the answers. Both answers are 5-digit numbers.
In June 2016 this puzzle was presented as a challenge with no help provided inviting anyone who wished to submit a solution. Participation was below the previous months high mark but still third most in the series, with 55 persons giving it a try, of which only the 15 listed found the correct answer to the main question. (I decided to cut the listing there since none of the misses were close.)
Dont mess with Texas! Congratulations to Dan Baker, who was the first of only four solvers to come up with the correct answer to the more difficult tiebreaker. This is Dans first win, though hes been a steady contender, populating most of my leaderboards. The other three perfect solvers were certainly no surprise, as adding up their combined wins might be as complicated as the problem.
|2||Dan Gheorghiu||British Columbia||47340||48360|
|6||Ivan Loy||Hong Kong||47340||48444|
|12||Wayne Somerville||Northern Ireland||47340||47808|
For the main question, one fact should be clear: Any hand cold for 12 tricks in a suit must have two voids, as otherwise it would be an easy setup for the defenders to crossruff two side suits. With that in mind, listing the possible hand types is straightforward, as shown in the table below. A question mark (?) means any card below the lowest specified card (or in Case 1 any card, period); the x in Case 2 means any card but the ace. All that remains is to determine the number of combinations for each case and multiply by the number of ways, which is always 12, since any hand with two voids has 12 suit permutations. Adding up the hands for each case gives the answer of 47,340.
|2||AK?????????? x||11c10 × 12c1||12||1584|
|3||AK????????? A?||11c9 × 12c1||12||7920|
|4||AKQ??????? AK?||10c7 × 11c1||12||15840|
|5||AKQJ????? AKQ?||9c5 × 10c1||12||15120|
|6||AKQJT??? AKQJ?||8c3 × 9c1||12||6048|
|7||AKQJT9? AKQJT?||7c1 × 8c1||12||672|
|Total hands guaranteed to win 12 tricks in a suit||47340|
With 12 trumps and a side ace (Case 1) trump quality makes no difference, since the most you could ever lose is one trump trick; but with a side loser (Case 2) they must be headed by A-K to withstand a potential trick-one loss and trump promotion. In Cases 3-7, trumps need only be strong enough to ensure drawing the missing trumps, since an opening ruff spends a trump, allowing you to ruff high on the return. Also note that the side suit does not have to be solid like Timothys A-K-Q-J-10; the lowest card can be a loser, since it vanishes with an opening ruff, and otherwise will be the only trick lost.
Dan Baker: The hand must have two voids (11-1-1-0 could suffer a crossruff); 7-6 must be AKQJT9? AKQJT? (7 × 8), [Dan accounts similarly for all other shapes]; then multiply by 12 for suit identity.
Jamie Pearson: Declarer must have exactly two suits (a third could lead to two ruffs). There must be enough top trumps not to allow an opening ruff and trump promotion, and at most one non-winner in the other suit. From there its just crunching the numbers and dealing with the special case of 12 trumps.
Sherman Yuen: Declarer must have exactly one side suit. If declarer has 7-11 trumps, the defense can always start with a ruff, followed by leading declarers void for a trump promotion; so the minimum holding is the top N trumps and the top N-1 cards in the side suit, where N is the number of missing trumps. If declarer has 12 trumps, they can be headed by AK with any singleton, or AQ/KQ with a singleton ace.
Gareth Birdsall: The hand must have only two suits, else concede a crossruff, and the lowest card in the side suit may be a loser. Total is: 12 × (2 + 11×13 + 11c9×12 + 10c7×11 + 9c5×10 + 8c3×9 + 7c1×8). [Gareth accounts for my Cases 1-2 differently, as did Sherman above, but its just as correct].
Dan Gheorghiu: A straightforward calculation. The hand cannot contain more than two suits for the reason explained in the preamble. Various lengths of the two suits, starting with 7-6 and ending with 12-1, yield a bell-shaped distribution, which peaks at 9-4 and 10-3 with almost the same number of hands.
The tiebreaker for hands guaranteed to make 6 NT was more difficult, as many more possibilities exist. Besides hands with 12 top tricks, one must account for hands able to establish a twelfth trick. My approach was to break it down into 51 cases listed in the following table, which is ordered by the length, secondarily by the strength, of the suit with a potential loser. Notations used: H = any card above the highest specified card; ? = any card below the lowest specified card; x = any card below and not touching the lowest specified card. Tinted rows indicate cases where the twelfth trick must be established, mandating at least two cards (A-K) in each solid suit.
|1||AKQJTxx AK AK AK||7c2||4||84|
|2||HHHH98? AK AK AK||5c4 × 6c1||4||120|
|3||AKQJTx AKQJT A A||7c1||12||84|
|4||AKQJTx AKQJ AK A||7c1||24||168|
|5||AKQJTx AKQ AKQ A||7c1||12||84|
|6||AKQJTx AKQ AK AK||7c1||12||84|
|7||HHHH98 AKQ AK AK||5c4||12||60|
|8||AKQJx AKQJT9 A A||8c1||12||96|
|9||AKQJx AKQJT AK A||8c1||24||192|
|10||AKQJx AKQJ AKQ A||8c1||24||192|
|11||AKQJx AKQJ AK AK||8c1||12||96|
|12||AKQJx AKQ AKQ AK||8c1||12||96|
|13||HHHT9 AKQJ AK AK||4c3||12||48|
|14||HHHT9 AKQ AKQ AK||4c3||12||48|
|15||AKQx AKQJT9? A A||9c1 × 7c1||12||756|
|16||AKQx AKQJT9 AK A||9c1||24||216|
|17||AKQx AKQJT AKQ A||9c1||24||216|
|18||AKQx AKQJT AK AK||9c1||12||108|
|19||AKQx AKQJ AKQJ A||9c1||12||108|
|20||AKQx AKQJ AKQ AK||9c1||24||216|
|21||AKQx AKQ AKQ AKQ||9c1||4||36|
|22||HHJT AKQJT AK AK||3c2||12||36|
|23||HHJT AKQJ AKQ AK||3c2||24||72|
|24||HHJT AKQ AKQ AKQ||3c2||4||12|
|25||AKx AKQJT??? A A||10c1 × 8c3||12||6720|
|26||AKx AKQJT9? AK A||10c1 × 7c1||24||1680|
|27||AKx AKQJT9 AKQ A||10c1||24||240|
|28||AKx AKQJT9 AK AK||10c1||12||120|
|29||AKx AKQJT AKQJ A||10c1||24||240|
|30||AKx AKQJT AKQ AK||10c1||24||240|
|31||AKx AKQJ AKQJ AK||10c1||12||120|
|32||AKx AKQJ AKQ AKQ||10c1||12||120|
|33||HQJ AKQJT9 AK AK||2c1||12||24|
|34||HQJ AKQJT AKQ AK||2c1||24||48|
|35||HQJ AKQJ AKQJ AK||2c1||12||24|
|36||HQJ AKQJ AKQ AKQ||2c1||12||24|
|37||Ax AKQJ????? A A||11c1 × 9c5||12||16632|
|38||Ax AKQJT??? AK A||11c1 × 8c3||24||14784|
|39||Ax AKQJT9? AKQ A||11c1 × 7c1||24||1848|
|40||Ax AKQJT9? AK AK||11c1 × 7c1||12||924|
|41||Ax AKQJT9 AKQJ A||11c1||24||264|
|42||Ax AKQJT9 AKQ AK||11c1||24||264|
|43||Ax AKQJT AKQJT A||11c1||12||132|
|44||Ax AKQJT AKQJ AK||11c1||24||264|
|45||Ax AKQJT AKQ AKQ||11c1||12||132|
|46||Ax AKQJ AKQJ AKQ||11c1||12||132|
|47||KQ AKQJT9? AK AK||7c1||12||84|
|48||KQ AKQJT9 AKQ AK||1||24||24|
|49||KQ AKQJT AKQJ AK||1||24||24|
|50||KQ AKQJT AKQ AKQ||1||12||12|
|51||KQ AKQJ AKQJ AKQ||1||12||12|
|Total hands guaranteed to win 12 tricks in notrump||48360|
Curiously, the most common hand that will guarantee 6 NT contains a nine-card suit (Case 37). This is also the longest possible suit, because a 10-carder would require three aces on the side, so the hand would either be cold for 13 tricks (excluded by the problem conditions) or have no guarantee of 12 after an ace is knocked out.
Dan Baker: With no singleton, there are 12 choices for an imperfect suit fewer than seven cards, e.g., five cards can be AKQJx without the 10 (8 choices) or HHHT9 (4 choices); or 51 choices if seven cards (7-2-2-2): AKQJTxx without the nine (21 choices) or HHHH98? (30 choices). With a singleton, you cant establish slow tricks after that ace is knocked out, so the imperfect suit must be 2-6 cards with 13-L choices (L = length), e.g., three cards must be AKx without the queen (13 - 3 = 10 choices).
Dan Gheorghiu: Declarer must have one suit with [the lowest card] a loser (e.g, AKQx) or a sequence broken by the absence of only one card (e.g., AKJT or KQJT). In the latter case, at least doubleton stoppers (AK) are necessary in every other suit, [because the lead must be surrendered to establish one or more tricks].
With some hands 6 NT is always makable but not guaranteed. For example, holding A-K-Q-J-9-2 A-K-Q A-K A-K, after any safe lead declarer can test spades to discover they cant be established, and without risk come down to J-9-2 A A A. The opponent must still have 10-8-x (else just concede a spade) so an endplay looms if declarer can strip his remaining cards, but theres the rub. Which ace do you cash first? If you pick a suit in which your opponent has two cards, you fail. And if the first ace cashed is successful, you have the same problem on the next one. Therefore, these hands require guessing the layout, which the problem conditions explicitly forbade.
Another curiosity, as well as my inspiration for this puzzle, is how close the two answers (47,340 and 48,360) happen to be. One could argue that a difference of 1020 is sizable, but in a playing field of 635+ billion bridge hands 635,013,559,600 to be exact its pretty darn close.
In case anyone is interested, the number of hands guaranteed to make 7 NT is exactly 3756 an easier problem solved many years ago. A summary table can be seen here.
Dan Gheorghiu: Evidently the Professor was right all along with his boson deflection theory, as hands with the highest freakness (eight- or nine-card suit) account for about 75 percent of those guaranteed to make 6 NT. Rumors are he will be nominated for a Fundamental Physics Prize for applying the laws of physics to bridge. My sincere congratulations, sir!
Grant Peacock: I spent the afternoon counting the number of ways to put a loser into a suit.
Well, so did my wife 20 years ago when she took me to a menswear store.
© 2016 Richard Pavlicek